Acapacitor is charged to a potential of 12.0 v and is then connected to a voltmeter having an internal resistance of 3.60 mω . after a time of 3.90 s the voltmeter reads 2.8 v . part a what is the capacitance?

As we know that when a charged capacitor is connected across a resistor then its voltage will decreased

This is known as discharging of capacitor

as we know that the equation of discharging is given as

here we know that

t = 3.90 s

now from above equation we have

as we know that

so capacitance is 0.75 micro farad

step-by-step explanation:

we are given

we have to simplify it

firstly, we will factor it

now, we can cancel like terms

now, we can combine it

so, we get

i think the answer would be b